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Prove that $\mathbb{Z}_5[x]$ is a unique factorization domain.

My approach is to prove that $\mathbb{Z}_5[x]$ is a PID, which implies that it is a UFD.

Proof:

Suppose there exists an ideal $I$ in $\mathbb{Z}_5[x]$ such that it is generated by two or more elements of $\mathbb{Z}_5[x]$. That is, $I = \langle g_1(x), g_2(x), ..., g_n(x)\rangle$. Then $I=\{a_1(x)g_1(x)+a_2(x)g_2(x)+...+a_n(x)g_n(x):a_i(x)\in \mathbb{Z}_5[x] \}$. Consider $\max\{a_i(x)g_i(x)\}=\deg_{max} (I)$. Then, since $\mathbb{Z}_5$ is a PID, $\langle a_i(x)g_i(x)\rangle = \langle g_1(x), g_2(x), ..., g_n(x)\rangle$. Hence, $\mathbb{Z}_5[x]$ is a PID. This implies that $\mathbb{Z}_5[x]$ is a UFD.

It would be interesting to know one's opinion on my proof.

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    Presumably you mean that $\mathbb Z_5[x]$ is a PID, not $\mathbb Z_5$. – 2017-02-27
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    I don't understand how $I$ is defined: on the right side of the colon, in the set builder notation, on expects some *property*? You don't have a property, but a combination of polynomials, i.e. a mathematical object. In other words, your definition is meaningless. – 2017-02-27
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    You are supposing that every ideal is finitely generated. It's an easier route to use the well-ordering principle followed by the fact that there is a division algorithm for polynomials over a field. – 2017-02-27
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    @Bernard Thanks, corrected that. – 2017-02-27
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    And what is $\deg_{\max}(I)$? – 2017-02-27
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    @Bernard The maximal degree of a summand in the generator of $I$. – 2017-02-27
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    This is very confusing… What is a summand in the generator (s, I guess)? – 2017-02-27
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    Sorry, I mean the maximal degree of a term in the sum in the generator of $I$, presented by the set above. – 2017-02-27
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    @Oiler Can you please elaborate? – 2017-02-27
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    @sequence if you have a proper, nontrivial ideal then there exists a polynomial of minimal degree. Show that minimal polynomial generates the entire ideal by applying the division algorithm for polynomials. – 2017-02-27

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Hint:

You have to use that, in a polynomial ring over a field, you can perform Euclidean divisions, and consider a non-zero polynomial of least degree in the ideal.

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    Do you mean the idea is to show that if there is a proper ideal $I$, which contains a polynomial $f(x)$ of minimum degree, then if we divide any polynomial in $\mathbb{Z}_5[x]$ by $f(x)$, we will get $g(x) = f(x)q(x)+r(x)$, which is any polynomial in $\mathbb{Z}_5[x]$, and hence $I$ cannot be a proper ideal? – 2017-02-27
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    You have to show that if $g(x)\in I$, then $r(x)=0$, henace any polynomial in $I$ is divisible by $f(x)$. Same argument as for showing $\mathbf Z$ (actually, all *Euclidean domains* are PIDs). – 2017-02-27
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By $\mathbb{Z}_5$ I assume you mean $\mathbb{Z}/5\mathbb{Z}$. This is a field, since 5 is prime. Any field is a PID. See Why any field is a principal ideal domain?

EDIT: After comments, I realized that while the OP said, "My approach is to prove that $\mathbb{Z}_5$ is a PID," what the OP meant to say was "My approach is to prove that $\mathbb{Z}_5[x]$ is a PID." My answer, while a good answer to the first, is not a good answer to the second.

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    But they are asking about the polynomial ring over $\mathbb{Z}/5\mathbb{Z}$, which is not a field. – 2017-02-27
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    I assumed that the OP was trying to show that $\mathbb{Z}/5\mathbb{Z}$ is a PID, as is indicated in the post, but as @ThomasAndrews points out in the comment above, I may have misread. – 2017-02-27
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    @Oiler: But if $R$ is an UFD, then so is $R[X]$. – 2017-02-27
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    @JoshuaRuiter Then I would need to prove that if $\mathbb{Z}_5$ is a PID then $\mathbb{Z}_5[x]$ is a PID. – 2017-02-27
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    @HenningMakholm While true, it doesn't seem to be the route this answer is suggesting and is kind of a "big theorem" for this exercise. – 2017-02-27
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    @sequence: No, a polynomial ring over a PID which is not a field is never a PID. – 2017-02-27
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    @Bernard We know that $\mathbb{Z}_5$ is a field. – 2017-02-27
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    @JoshuaRuiter Sorry that was a typo, fixed now. – 2017-02-27
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    @sequence: That's right, but the reason you invoked was only it's a PID, which is not a correct argument ($\mathbf Z[x]$ is *not* a PID). – 2017-02-27